### Table 4: Average number of Newton iterations for solving the linearized model in non-linear Fair-estimation.

in Solution of Linear Programming and Non-Linear Regression Problems Using Linear M-Estimation Methods

1999

"... In PAGE 109: ...Table4 : Results for the updating routine of the software package when used as a tool for nding L from scratch. Times are given in seconds.... ..."

### Table 1. Results: nonlinear problems.

2007

Cited by 2

### Table 1. Results: nonlinear problems.

2007

Cited by 2

### Table 1. pMSE and Bias of Power Estimates for Two-sample t-Statistic with Parametric Bootstrap Critical Values O = 1000, I = 59

2000

"... In PAGE 12: ... We look at two (O; I) combinations, (O = 1000; I = 59) and (O = 596; I = 99), that have about 59000 computations each. Table1 reports estimates of the root mean squared error (pMSE) and bias 1000 of the various power estimates. The standard errors of the estimates are in the range .... In PAGE 12: ...002 for pMSE and around 2 for the bias 1000. The rst and seventh rows (p1) of Table1 give results for power estimates based on the true known t percentiles appropriate for normal data. They are labeled p1 to re ect the fact that resampling with I approaching 1 would give this result.... In PAGE 12: ... They are labeled p1 to re ect the fact that resampling with I approaching 1 would give this result. These of course are unbiased (the nonzero bias results in Table1 just re ect Monte Carlo variation), and here pMSE could have been calculated simply by ppower(1-power)=O. For a given O, p1 represents the best power estimates possible.... In PAGE 12: ... For these raw estimates the (O = 596; I = 99) situation is more e cient in terms of pMSE than (O = 1000; I = 59) for all but = 0:5 because the bias is a large factor except at = 0:5. The other estimators in Table1 are 1. b plin: the simple linear extrapolation method using (5) for the (O = 1000; I = 59) case and (6) for the (O = 596; I = 99) case.... In PAGE 13: ...a;bb) distribution. From Table1 we see that the the linear extrapolation estimators, b plin and b pgls, perform the best and very similarly. Their similarity is likely due to the fact (not displayed) that the estimated covariance matrix of the b pI used as dependent variables in the regressions has nearly equal diagonal elements and nearly equal o -diagonal elements.... ..."

Cited by 3

### Table 11. Timings (in seconds) for nonlinear problem

1995

Cited by 6

### Table 1. Solutions of nonlinear problem. Mixed Integer Continuous

"... In PAGE 9: ... Table1 shows that OptQuest finds the best solution when the problem contains discrete variables (with an objective value that is 3.... ..."

### Table 2 Maximum error for the nonlinear problem by Algorithm 1.

61

"... In PAGE 6: ...3) to test Algorithm 1. Figure 4 and Table2 show the computational results with ne mesh hx = hy = 1 100, and coarse mesh Hx = Hy = 1... ..."

### Table 1: Results for nonlinear complementarity problems

"... In PAGE 12: ... If in the rst iteration the stepsize of one is accepted we use the nomonotone strategy of Algorithm B, otherwise we use algorithm A for the rst ve iterations and then switch to Algorithm B. In all the algorithms the following constants were used: quot; = 1; = 10?4; = 10?8; p = 2:1; = 0:9: We tried the algorithm on several test problems taken from the literature, we considered nonlinear ( Table1 ) but also linear problems (Table 2); the test set includes problems that are not P0 and problems that are not R-regular or b-regular at the solution. Some details on these problems, the starting points, and adequate references are reported in the appendix.... In PAGE 15: ... This problem has two solutions: x1 = (0:5p6; 0; 0; 0:5) and x2 = (1; 0; 3; 0); x1 fails to be R-regular and is degenerate. In the results of Table1 convergence always occurred to x1 except in the case indicated by an asterisk. The linearized complementarity problem at 0 has no solution.... ..."

### Table 12: Results for Interpolated Nonlinear CSTR Problem

"... In PAGE 25: ...1 ; zi?1 2 ; ui) = (zi 1; zi 2; zi 3) for all i = 1; : : : ; N z0 1 = 0:09 z0 2 = 0:09 Results for convergence to an absolute tolerance of 5 10?3 are presented in Table12 . Using these values as a starting point, the local solver obtained the solutions in Table 13.... ..."